# Converting integration dV in spherical coordinates for volume but not for surface?

When calculating the volume of a spherical solid, i.e. a triple integral over angles and radius, the standard $$dx\,dy\,dz$$ gets converted into $$f(x,y,z)r^2\sin\Phi \,d\Phi \,d\Theta \,dr$$. However, it seems that when we calculate a spherical surface integral, that is not the case, and we instead just have $$f(x,y,z)\left|\frac{\delta r}{\delta \Phi}\times\frac{\delta r}{\delta \Theta}\right|\,d\Phi \,d\Theta$$.

Why is that? I'm just confused about when I should "convert" when parametrizing a surface and when not.

The short answer to your question is that spherical coordinates only refers to the the three variable coordinate transformation. A parameterization should never be confused with a coordinate system, especially when that distinction is critical for doing calculus.

However we do have an interesting formula for certain kinds of surfaces derived as constants in some other coordinate system. Let's say you have some new (orthogonal*, very important) coordinate system

$$\vec{r}(u,v,w) = (x(u,v,w),y(u,v,w),z(u,v,w))$$

with a known Jacobian $$J(u,v,w)$$. Then if you were to parameterize a surface by

$$\vec{r_s}(v,w) = \vec{r}(u=k,v,w)$$

then we have the following formula:

$$\left|\frac{\partial \vec{r_s}}{\partial v}\times\frac{\partial \vec{r_s}}{\partial w}\right| = \frac{J}{\left|\frac{\partial \vec{r}}{\partial u}\right|}$$

$$f(x,y,z)r^2\sinΦ\ dΦ\ dΘ\ dr$$ gets converted to $$R^2\ f(x,y,z) \ \sinΦ \ dΦ\ dΘ$$ where $$R$$ is the radius of the sphere.

To convert to spherical coordinates rewrite the differential form of volume multiped by the Jacobian of coordinate transformation matrix after evaluation

$$\frac{\partial (x,y,z)}{\partial (r,\Theta, \Phi)}=r^2 \sin \Phi$$

changing triple integrals into $$r^2\sin\Phi \,d\Phi \,d\Theta \,dr$$

For surface integrals evaluate (2X2) matrix $$\left|\frac{\delta r}{\delta \Phi}\times\frac{\delta r}{\delta \Theta}\right|\,d\Phi \,d\Theta$$.